Question

0, (6)x+2x-0, (3)=4-2 (x+1)+3, (6)

Answer 1

   
$0,(6)x+2x-0,(3)=4-2(x+1)+3,(6) \\ \\ \frac{6}{9} x+2x-\frac{3}{9} =4-2x-2+ \frac{36-3}{9} \\ \\ \frac{2}{3} x+2x-\frac{1}{3} =4-2x-2+ \frac{33}{9} \\ \\ \frac{2}{3} x+2x-\frac{1}{3} =4-2x-2+ \frac{11}{3} \;\;\;\;\; (Aducem\;la\;acelasi\;numitor) \\ \\ 2x+6x-1=12-6x-6+11 \\ 2x+6x+6x=12-6+11+1 \\ 14x=18 \\ \\ x= \frac{18}{14} =\boxed{ \frac{9}{7} }=\boxed{1,(285714)}$

Answer 2

0,(6)x+2x-0,(3)=4-2(x+1)+3,(6)
6x +2x- 3 =4-2x- 2+36-3
9         9                 9

6x +2x- 3 =4-2x- 2+ 33     ADUC LA ACELAŞI NUMITOR!
9         9                 9

6x+2·9x- 3=4·9-2·9x-2·9+33
6x+18x- 3=36-18x-18+33
24x- 3=18-18x+33
24x- 3=51-18x
24x+18x=51+ 3
42x=54 :6
  7x=9
    x=  9
        7